The bounty period lasts 7 days. Bounties must have a minimum duration of at least 1 day. After the bounty ends, there is a grace period of 24 hours to manually award the bounty. Simply click the bounty award icon next to each answer to permanently award your bounty to the answerer. (You cannot award a bounty to your own answer.)

@Mathphile I found no prime of the form $$n^{n+1}+(n+1)^{n+2}$$ for $n>392$ yet and neither a reason why the expression cannot be prime for odd n, although there are far more even cases without a known factor than odd cases.

@TheSimpliFire That´s what I´m thinking about, I had some "vague feeling" that there must be some elementary proof, so I decided to find it, and then I found it, it is really "too elementary", but I like surprises, if they´re good.

It is in fact difficult, I did not understand all the details either. But the ECM-method is analogue to the p-1-method which works well, then there is a factor p such that p-1 is smooth (has only small prime factors)

Brocard's problem is a problem in mathematics that asks to find integer values of n and m for which
n
!
+
1
=
m
2
,
{\displaystyle n!+1=m^{2},}
where n! is the factorial. It was posed by Henri Brocard in a pair of articles in 1876 and 1885, and independently in 1913 by Srinivasa Ramanujan.
== Brown numbers ==
Pairs of the numbers (n, m) that solve Brocard's problem are called Brown numbers. There are only three known pairs of Brown numbers:
(4,5), (5,11...

$\textbf{Corollary.}$ No solutions to Brocard's problem (with $n>10$) occur when $n$ that satisfies either \begin{equation}n!=[2\cdot 5^{2^k}-1\pmod{10^k}]^2-1\end{equation} or \begin{equation}n!=[2\cdot 16^{5^k}-1\pmod{10^k}]^2-1\end{equation} for a positive integer $k$. These are the OEIS sequences A224473 and A224474.

Proof: First, note that since $(10^k\pm1)^2-1\equiv((-1)^k\pm1)^2-1\equiv1\pm2(-1)^k\not\equiv0\pmod{11}$, $m\ne 10^k\pm1$ for $n>10$. If $k$ denotes the number of trailing zeros of $n!$, Legendre's formula implies that \begin{equation}k=\min\left\{\sum_{i=1}^\infty\left\lfloor\frac n{2^i}\right\rfloor,\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\right\}=\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\end{equation} where $\lfloor\cdot\rfloor$ denotes the floor function.

The upper limit can be replaced by $\lfloor\log_5n\rfloor$ since for $i>\lfloor\log_5n\rfloor$, $\left\lfloor\frac n{5^i}\right\rfloor=0$. An upper bound can be found using geometric series and the fact that $\lfloor x\rfloor\le x$: \begin{equation}k=\sum_{i=1}^{\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor\le\sum_{i=1}^{\lfloor\log_5n\rfloor}\frac n{5^i}=\frac n4\left(1-\frac1{5^{\lfloor\log_5n\rfloor}}\right)<\frac n4.\end{equation}

Thus $n!$ has $k$ zeroes for some $n\in(4k,\infty)$. Since $m=2\cdot5^{2^k}-1\pmod{10^k}$ and $2\cdot16^{5^k}-1\pmod{10^k}$ has at most $k$ digits, $m^2-1$ has only at most $2k$ digits by the conditions in the Corollary. The Corollary if $n!$ has more than $2k$ digits for $n>10$. From equation $(4)$, $n!$ has at least the same number of digits as $(4k)!$. Stirling's formula implies that \begin{equation}(4k)!>\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\end{equation}

Since the number of digits of an integer $t$ is $1+\lfloor\log t\rfloor$ where $\log$ denotes the logarithm in base $10$, the number of digits of $n!$ is at least \begin{equation}1+\left\lfloor\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)\right\rfloor\ge\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right).\end{equation}

Therefore it suffices to show that for $k\ge2$ (since $n>10$ and $k<n/4$), \begin{equation}\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)>2k\iff8\pi k\left(\frac{4k}e\right)^{8k}>10^{4k}\end{equation} which holds if and only if \begin{equation}\left(\frac{10}{\left(\frac{4k}e\right)}\right)^{4k}<8\pi k\iff k^2(8\pi k)^{\frac1{4k}}>\frac58e^2.\end{equation}

Now consider the function $f(x)=x^2(8\pi x)^{\frac1{4x}}$ over the domain $\Bbb R^+$, which is clearly positive there. Then after considerable algebra it is found that \begin{align*}f'(x)&=2x(8\pi x)^{\frac1{4x}}+\frac14(8\pi x)^{\frac1{4x}}(1-\ln(8\pi x))\\\implies f'(x)&=\frac{2f(x)}{x^2}\left(x-\frac18\ln(8\pi x)\right)>0\end{align*} for $x>0$ as $\min\{x-\frac18\ln(8\pi x)\}>0$ in the domain.

Thus $f$ is monotonically increasing in $(0,\infty)$, and since $2^2(8\pi\cdot2)^{\frac18}>\frac58e^2$, the inequality in equation $(8)$ holds. This means that the number of digits of $n!$ exceeds $2k$, proving the Corollary. $\square$

We get $n^n+3\equiv 0\pmod 4$ for odd $n$, so we can see from here that it is even (or, we could have used @TheSimpliFire's one-or-two-step method to derive this without any contradiction - which is better)

@TheSimpliFire Hey! with $4\pmod {10}$ and $0\pmod 4$ then this is the same as $10m_1+4$ and $4m_2$. If we set them equal to each other, we have that $5m_1=2(m_2-m_1)$ which means $m_1$ is even. We get $4\pmod {20}$ now :P

Yet again a conjecture!
Motivated by Catalan's conjecture and a recent question of mine, I conjecture that
For distinct, positive integers $a,b$, the only solution to this equation $$a^b-b^a=a+b\tag1$$ is $(a,b)=(2,5).$
It is of anticipation that there will be much fewer solutions for incr...